Two infinite length wires carry currents 8 A | vedclass

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(D) The magnetic field due to an infinitely long wire carrying current $I$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.F

Vedclass · Accepted Answer

$\frac{5 \mu_0}{\pi d}$

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(D) The magnetic field due to an infinitely long wire carrying current $I$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.For the wire along the $X$-axis carrying current $I_1 = 6 \ A$,the magnetic field at $P(0, 0, d)$ is directed along the $Y$-axis (using the right-hand thumb rule). Thus,$\vec{B}_1 = \frac{\mu_0 I_1}{2 \pi d} \hat{j} = \frac{\mu_0 (6)}{2 \pi d} \hat{j} = \frac{3 \mu_0}{\pi d} \hat{j}$.For the wire along the $Y$-axis carrying current $I_2 = 8 \ A$,the magnetic field at $P(0, 0, d)$ is directed along the negative $X$-axis. Thus,$\vec{B}_2 = -\frac{\mu_0 I_2}{2 \pi d} \hat{i} = -\frac{\mu_0 (8)}{2 \pi d} \hat{i} = -\frac{4 \mu_0}{\pi d} \hat{i}$.The net magnetic field at $P$ is $\vec{B}_P = \vec{B}_1 + \vec{B}_2 = -\frac{4 \mu_0}{\pi d} \hat{i} + \frac{3 \mu_0}{\pi d} \hat{j}$.The magnitude of the magnetic field is $B_P = \sqrt{\left(-\frac{4 \mu_0}{\pi d}\right)^2 + \left(\frac{3 \mu_0}{\pi d}\right)^2} = \frac{\mu_0}{\pi d} \sqrt{16 + 9} = \frac{5 \mu_0}{\pi d}$.

Two infinite length wires carry currents $8 \ A$ and $6 \ A$ respectively and are placed along the $X$ and $Y$ axes respectively. The magnetic field at a point $P(0, 0, d)$ will be:

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